Suppose we have a function that is analytic on the open unit disk.

Suppose we have a continuous function on the boundary of the disk that maps each point on the boundary of the disk to its conjugate.

Suppose that near the boundary of the unit disk, the values of our analytic function approach the values of our continuous function.

Can I say that the analytic function has to be identically the conjugation function? I'm looking find a contradiction like this. (In this case, I know that the conjugation function is not analytic.)

Complex analysis isn't my strongest subject. I'll appreciate any help you can give me!


1 Answer 1


Yes, the reasoning you indicated works. Let $\mathbb D$ be the open unit disk. Recall the maximum principle for harmonic functions: if $u:\mathbb D\to \mathbb R$ is harmonic, and $u(z_n)\to 0$ for any sequence $z_n$ such that $|z_n|\to 1$, then $u\equiv 0$.

This principle can be applied to complex-valued functions by considering their real part and imaginary parts separately. If $f$ is the analytic function you described, and $g(z)=\bar z$ is the conjugation, then both $\operatorname{Re}(f-g)$ and $\operatorname{Im}(f-g)$ are harmonic, with zero boundary values. Therefore, $f\equiv g$.

More generally: if $f$ is analytic, $g$ is antianalytic, and $f-g$ vanishes on the boundary, then $f \equiv g $, which implies $f$ is a constant function.

Another approach is available in your case: as $r\to 1$, $$ 0 = \int_{|z|=r} f(z)\,dz \to \int_{|z|=1} \bar z\,dz = \int_{|z|=1} z^{-1}\,dz =2\pi i $$ a contradiction.

  • $\begingroup$ Thank you for explaining it so clearly! I hadn't thought to apply the maximum principle for harmonic functions here; great idea! $\endgroup$ Nov 9, 2013 at 22:20
  • $\begingroup$ @John You're welcome. For your information, the checkmark next to answers can be used to indicate that the answer resolved the issue (if this is indeed so). See meta.math.stackexchange.com/a/3287/103402 $\endgroup$
    – user103402
    Nov 9, 2013 at 22:25
  • $\begingroup$ Done. Thanks again. $\endgroup$ Nov 9, 2013 at 22:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .